sum_{k=0}^4 sin^2 left( (2k+1) frac{pi}{20} r | vedclass

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(B) ધારો કે $S = \sum_{k=0}^4 \sin^2 \left( (2k+1) \frac{\pi}{20} \right)$.સરવાળાને વિસ્તૃત કરતા:$S = \sin^2 \frac{\pi}{20} + \sin^2 \frac{3\pi}{20} +

Vedclass · Accepted Answer

$\frac{5}{2}$

Vedclass · Answer

(B) ધારો કે $S = \sum_{k=0}^4 \sin^2 \left( (2k+1) \frac{\pi}{20} ight)$.સરવાળાને વિસ્તૃત કરતા:$S = \sin^2 \frac{\pi}{20} + \sin^2 \frac{3\pi}{20} + \sin^2 \frac{5\pi}{20} + \sin^2 \frac{7\pi}{20} + \sin^2 \frac{9\pi}{20}$.નિત્યસમ $\sin^2 	heta = \cos^2 \left( \frac{\pi}{2} - 	heta ight)$ નો ઉપયોગ કરતા:$\sin^2 \frac{\pi}{20} = \cos^2 \frac{9\pi}{20}$ અને $\sin^2 \frac{3\pi}{20} = \cos^2 \frac{7\pi}{20}$.તેથી,$S = (\sin^2 \frac{9\pi}{20} + \cos^2 \frac{9\pi}{20}) + (\sin^2 \frac{7\pi}{20} + \cos^2 \frac{7\pi}{20}) + \sin^2 \frac{\pi}{4}$.$S = 1 + 1 + \frac{1}{2} = \frac{5}{2}$.

$\sum_{k=0}^4 \sin^2 \left( (2k+1) \frac{\pi}{20} \right) =$

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